Functional Analysis: when is addition/multiplication of a closure not the same as the closure of the addition/multiplication? The continuity of + and · operators imply that $\lambda\overline{A}=\overline{\lambda A}$ 
and $\overline{A}+\overline{B} \subseteq \overline{ A + B}$
Are there cases where equality does not necessarily hold?
Issue: all the sequences in A,B that I come up with seem to satisfy the above relations, but I was told it need not hold. 
 A: First of all, we must have equality between $\lambda \overline{A}$ and $\overline{\lambda A}$. Continuity of scalar multiplication implies that
$$\lambda \overline{A} \subseteq \overline{\lambda A}.$$
If $\lambda \neq 0$, then multiplying both sides by $\lambda^{-1}$,
$$\overline{A} \subseteq \lambda^{-1}\overline{\lambda A} \subseteq \overline{\lambda^{-1}\lambda A} = \overline{A},$$
hence
$$\overline{A} = \lambda^{-1}\overline{\lambda A} \implies \lambda \overline{A} = \overline{\lambda A}.$$
When $\lambda = 0$, the equality clearly holds, as both sides are equal to $\{0\}$.
Now, the corresponding equality for summation need not hold. In order to figure out an example, it helps to know that if $\overline{A}$ or $\overline{B}$ are compact, then equality does indeed hold. To prove this, suppose $x_n \in A + B$ converges to some $x$. Then there exist sequences $a_n$ in $A$ and $b_n$ in $B$ such that $x_n = a_n + b_n$. If $\overline{A}$ or $\overline{B}$ is compact, then we can simply take a subsequence $n_k$ so that $a_{n_k}$ or $b_{n_k}$ converges. If $a_{n_k} \to a \in \overline{A}$, then
$$b_{n_k} = x_{n_k} - a_{n_k} \to x - a \in \overline{B},$$
hence $x = a + (x - a) \in \overline{A} + \overline{B}$. Similar logic works when $b_{n_k}$ converges to some $b \in \overline{B}$.
So, in order to get a counterexample, you'll need to consider the sum of two sequences, in $A$ and $B$ respectively, each without any convergent subsequence. Here's the classic example:
Consider the space $\Bbb{R}^2$ and let
\begin{align*}
A &= \{(x, y) : x > 0 \text{ and }y \ge 1/x\} \\
B &= \{(x, y) : x < 0 \text{ and }y \ge -1/x\}.
\end{align*}
Note that $A$ and $B$ are closed. Also, we have $a_n = (n, 1/n) \in A$ and $b_n = (-n, 1/n) \in B$, but
$$a_n + b_n = (n, 1/n) + (-n, 1/n) = (0, 2/n) \to (0, 0) \in \overline{A + B}.$$
However, $(0, 0) \notin \overline{A} + \overline{B} = A + B$, as every point in both $A$ and $B$ has a strictly positive $y$ coordinate. So, in this case, we have a strict subset, rather than equality.
A: Take for the space $V=\mathbb R^2$ endowed with the usual topology and
$$\begin{aligned}
A &= \{(x,y) \mid y \ge e^x\}\\
B &= \mathbb R \times \{0\}
\end{aligned}$$
Both $A,B$ are closed. However $A+B= \overline{A} + \overline{B} = \mathbb R \times (0,\infty)$ while $\overline{A+B} = \mathbb R \times [0,\infty)$.
